The value of $\frac{d}{d x}(|x-1|+|x-5|)$ at $x=3$ is

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

The value of

\frac{d}{d x}(|x-1|+|x-5|)

x=3

View answer and explanation

Correct option: B

Option B (0) is correct because at

x=3

, the function

f(x) = |x-1| + |x-5|

is constant, hence its derivative is zero.

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